A liquid that can be modeled as water of mass 0.25kg is heat to 80 degrees Celsius. The liquid is poured over ice of mass 0.070kg at 0 degrees Celsius. What is the temperature at thermal equilibrium, assuming no energy loss to the environment? How much energy must be removed from 0.085kg of steam at 120 degrees Celsius to form liquid water at 80 degrees Celsius?

Answers

Answer 1

Temperature at equilibrium is 0 degrees Celsius. Energy needed to remove from steam is 36.89 kJ.

1. At thermal equilibrium, the temperature of the liquid and ice mixture will be 0 degrees Celsius. To find the amount of energy required to reach thermal equilibrium, we use the equation:

Q = m * c * deltaT,

where

Q is the heat transferred,

m is the mass,

c is the specific heat capacity, and

deltaT is the change in temperature.

The heat transferred from the hot liquid to the ice is equal to the heat required to melt the ice and then raise its temperature to 0 degrees Celsius. Using this equation, we find that:

Q = 117.5 J.

2. To find the amount of energy that needs to be removed from the steam to form liquid water at 80 degrees Celsius, we use the equation:

Q = mL,

where

Q is the heat transferred,

m is the mass, and

L is the latent heat of vaporization.

First, we need to find the mass of the steam that needs to be condensed. We know that the total mass of the system is 0.085kg, so the mass of the steam can be found by subtracting the mass of the liquid water at 80 degrees Celsius from the total mass.

Using this equation, we find that the mass of the steam is 0.075kg. The latent heat of vaporization for water is 2.26 x [tex]10^6[/tex] J/kg.

Plugging in the values, we find that:

Q = 36.89 kJ.

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Answer 2

1a. The temperature at thermal equilibrium after pouring water (mass = 0.25 kg) at 80°C over ice (mass = 0.070 kg) at 0°C is approximately 0°C.

Determine the final temperature?

To find the final temperature at thermal equilibrium, we can apply the principle of conservation of energy. The heat lost by the water as it cools down will be equal to the heat gained by the ice as it melts.

The heat lost by the water can be calculated using the formula: Q₁ = m₁c₁ΔT₁, where m₁ is the mass of water, c₁ is the specific heat capacity of water, and ΔT₁ is the change in temperature.

The heat gained by the ice can be calculated using the formula: Q₂ = m₂L, where m₂ is the mass of ice and L is the latent heat of fusion.

At thermal equilibrium, Q₁ = Q₂. Therefore, m₁c₁ΔT₁ = m₂L.

Rearranging the equation, we have ΔT₁ = (m₂L) / (m₁c₁).

Substituting the given values, ΔT₁ = (0.070 kg * 334,000 J/kg) / (0.25 kg * 4,186 J/(kg·°C)) = 0.56 °C.

Since the initial temperature of the ice is 0°C, the final temperature at thermal equilibrium is approximately 0°C.

Note: The specific heat capacity of water (c₁) is 4,186 J/(kg·°C), and the latent heat of fusion (L) for ice is 334,000 J/kg.

1b. The amount of energy that must be removed from 0.085 kg of steam at 120°C to form liquid water at 80°C is approximately 244,400 J.

To find the energy?

To determine the energy that needs to be removed, we can calculate the heat lost by the steam as it cools down from 120°C to 80°C.

The heat lost by the steam can be calculated using the formula: Q = mcΔT, where m is the mass of steam, c is the specific heat capacity of steam, and ΔT is the change in temperature.

The specific heat capacity of steam (c) is approximately 2,010 J/(kg·°C).

Substituting the given values, Q = (0.085 kg * 2,010 J/(kg·°C)) * (120°C - 80°C) = 8,535 J/°C * 40°C = 341,400 J.

Therefore, the amount of energy that must be removed from 0.085 kg of steam at 120°C to form liquid water at 80°C is approximately 244,400 J.

Note: The specific heat capacity of steam (c) is approximate and may vary slightly with temperature.

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Complete question here:

1a. A liquid that can be modeled as water of mass 0.25kg is heated to 80 degrees celsius. The liquid is poured over ice of mass 0.070kg at 0 (zero) degrees celsius. What is the temperature at thermal equilibrium, assuming no energy loss to the environment?

1b. how much energy must be removed from 0.085kg of steam at 120 degrees celsius to form liquid water at 80 degrees celsius?


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Answers

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